Date: Fri, 7 Sep 2018 21:49:31 -0700 From: Steve Kargl <sgk@troutmask.apl.washington.edu> To: Bruce Evans <brde@optusnet.com.au> Cc: freebsd-numerics@freebsd.org Subject: Re: j0 (and y0) in the range 2 <= x < (p/2)*log(2) Message-ID: <20180908044931.GA52882@troutmask.apl.washington.edu> In-Reply-To: <20180907175406.GA49351@troutmask.apl.washington.edu> References: <20180903235724.GA95333@troutmask.apl.washington.edu> <20180905201540.D1142@besplex.bde.org> <20180905152104.GA26453@troutmask.apl.washington.edu> <20180906034525.A2959@besplex.bde.org> <20180905190223.GA27865@troutmask.apl.washington.edu> <20180906202658.K978@besplex.bde.org> <20180906223026.GA43005@troutmask.apl.washington.edu> <20180907230825.A933@besplex.bde.org> <20180907175406.GA49351@troutmask.apl.washington.edu>
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On Fri, Sep 07, 2018 at 10:54:06AM -0700, Steve Kargl wrote: > > I'll probably try a different approach tomorrow. j0(x), > cc, and ss are smoothly varying functions of x. u and v > appear to be slowly varying (in small intervals), so we can > have > > j0(x) = sqrt(1/pi) * (cc(x) *u-ss(x) *v) / sqrt(x). > j0(x+e) = sqrt(1/pi) * (cc(x+e)*u-ss(x+e)*v) / sqrt(x+e). > > where e is some small perturbution. Two equation and two > unknowns is easily solved. Simply need to do this across > the interval to generate u(x) and v(x). > I may have have a better approach! j0(x) = sqrt(1/pi) * (cc(x) * u - ss(x) * v) / sqrt(x) y0(x) = sqrt(1/pi) * (cc(x) * u + ss(x) * v) / sqrt(x) j0(x) + y0(x) = 2 * sqrt(1/pi) * cc(x) * u j0(x) - y0(x) = - 2 * sqrt(1/pi) * ss(x) * v So, we have u = (j0 + y0) / (2 * sqrt(1/pi) * cc) v = (y0 - j0) / (s * sqrt(1/pi) * ss) Thus, we can find u = r/s or u = 1 + r/s and v = r/s. -- Steve
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